Difference between revisions of "2017 AMC 10A Problems/Problem 4"
Carpedatdiem (talk | contribs) m (→Problem) |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Mia is | + | Mia is helping her mom pick up <math>30</math> toys that are strewn on the floor. Mia’s mom manages to put <math>3</math> toys into the toy box every <math>30</math> seconds, but each time immediately after those <math>30</math> seconds have elapsed, Mia takes <math>2</math> toys out of the box. How much time, in minutes, will it take Mia and her mom to put all <math>30</math> toys into the box for the first time? |
<math>\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5</math> | <math>\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5</math> |
Revision as of 18:19, 8 March 2017
Contents
Problem
Mia is helping her mom pick up toys that are strewn on the floor. Mia’s mom manages to put toys into the toy box every seconds, but each time immediately after those seconds have elapsed, Mia takes toys out of the box. How much time, in minutes, will it take Mia and her mom to put all toys into the box for the first time?
Solution
Every seconds toys are put in the box, so after seconds there will be toys in the box. Mia's mom will then put toys into to the box and we have our total amount of time to be seconds, which equals minutes.
Solution 2
Though Mia's mom places toys every seconds, Mia takes out toys right after. Therefore, after seconds, the two have collectively placed toy into the box. Thereforeby minutes, the two would have placed toys into the box. Therefore, at minutes, the two would have placed toys into the box. Though Mia may take toys out right after, the number of toys in the box first reaches by minutes.
See also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.